class Solution {
public:
    int minInsertions(string s) {
        int n  = s.size();
        vector<vector<int>> dp(n, vector<int>(n, 0));
        // dp[i][j] 表是字符串 [i, j] 区域成为回文子串的最少插入次数
        for(int i = n - 1; i >= 0; --i)
        {
            for(int j = i + 1; j < n; ++j)
            {
                if(s[i] == s[j])
                    dp[i][j] = dp[i + 1][j - 1];
                else
                    dp[i][j] = min(dp[i + 1][j], dp[i][j - 1]) + 1;
            }
        }
        return dp[0][n - 1];
    }
};
 